Time period of a particle executing `SHM` is `8` sec. At `t=0` it is at the mean position. The ratio of the distance covered by the particle in the `1st` second to the `2nd` second is:

Options

A. `(1)/(sqrt2+1)`
B. `sqrt2`
C. `(1)/(sqrt2)`
D. `sqrt2+1`

Answer

D

Explanation

Here `x=Asinomegat`
`x_1=Asin((2pi)/(8))xx1`
`=Asin((pi)/(4))=(A)/(sqrt2)`
and `x_2=Asin((2pi)/(8))xx2=A`
Therefore, the distance traveled in 2nd second is
`x_2'=x_2-x_1=A-(A)/(sqrt2)=((sqrt2-1)A)/(sqrt2)`
Ratio `=(x_1)/(x_2)=((A)/(sqrt2))/(((sqrt2-1)A)/(sqrt2))=(1)/((sqrt2-1))`
`=(sqrt2-1)/((sqrt2-1)(sqrt2+1))=((sqrt2-1))/(2-1)=(sqrt2-1)`